So certainly $f(x+1) \in \{f(x),f(x)+1\}$. On the other hand, every such function satisfies the condition. Suppose that $f(n)-f(1)=k$. Then we have $\binom{n-1}{k}$ choices of where the increments should be and $n-k$ choices of $f(1)$ yielding the result \[\sum_{k=0}^{n-1} (n-k) \binom{n-1}{k}=n2^{n-1}-(n-1)2^{n-2}\]